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Marina CMI [18]
3 years ago
5

Joel is using the two data points closest to the line to determine the equation of the regression line. What should Joel find fo

r the slope of the line? Round to three decimal places.

Mathematics
2 answers:
pav-90 [236]3 years ago
7 0
 3 4 is the answer  tos thi
vesna_86 [32]3 years ago
6 0
Slope=rise/run
we see that the slope will be negative
it is more accurate the farther we are apart
we wiill use the points closest to the line, those are (3.5,3.5) and (5,1)

slope between (x1,y1) and (x2,y2) is (y2-y1)/(x2-x1)
for this one
(1-3.5)/(5-3.5)=-2.5/1.5=-1.666666666≈-1.67

the slope is -1.67
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The intensity of an earthquake with a magnitude of 2 is 100 times greater than the intensity of an a standard earthquake .

<h3>What is magnitude of earthquake?</h3>

Magnitude of earthquake is the measure of the size of origin of the earthquake. The magnitude of the earthquake keeps the same value for each place.

An earthquake with a magnitude of about 2. 0 or less is called a micro-earthquake and not felt usually.The intensity of an earthquake with a magnitude of 2.

Let the intensity of this earthquake is <em>n </em>times greater than the intensity of an a standard earthquake. Thus the intensity of standard earthquake can be given as,

n=10^2\\n=100

If the magnitude would be 3 then the intensity would be,

n=10^3\\n=1000

It would be 1000 times greater than the standard earthquake and so on.

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3 years ago
The state education commission wants to estimate the fraction of tenth grade students that have reading skills at or below the e
marusya05 [52]

Answer:

The estimation for the proportion of tenth graders reading at or below the eighth grade level is given by:

\hat p =\frac{955-812}{955}= 0.150

0.150 - 1.64 \sqrt{\frac{0.150(1-0.150)}{955}}=0.131

0.150 + 1.64 \sqrt{\frac{0.150(1-0.150)}{955}}=0.169

And the 90% confidence interval would be given (0.131;0.169).

Step-by-step explanation:

We have the following info given:

n= 955 represent the sampel size slected

x = 812 number of students who read above the eighth grade level

The estimation for the proportion of tenth graders reading at or below the eighth grade level is given by:

\hat p =\frac{955-812}{955}= 0.150

The confidence interval for the proportion  would be given by this formula

\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}

For the 90% confidence interval the significance is \alpha=1-0.9=0.1 and \alpha/2=0.05, with that value we can find the quantile required for the interval in the normal standard distribution and we got.

z_{\alpha/2}=1.64

And replacing into the confidence interval formula we got:

0.150 - 1.64 \sqrt{\frac{0.150(1-0.150)}{955}}=0.131

0.150 + 1.64 \sqrt{\frac{0.150(1-0.150)}{955}}=0.169

And the 90% confidence interval would be given (0.131;0.169).

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